Walmart Company surveyed the senior managers in the retail industry to understand their mean annual salary....

Walmart Company surveyed the senior managers in the retail industry to understand their mean annual salary. A random sample of 49 managers reviled a sample mean of $45420. The standard deviation of this population is $2050. Calculate the 95% confidence interval estimate.

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Expert Solution

Solution :

Given that,

= 45420

= 2050

n = 49

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 1.96 * (2050 / $\sqrt$ 49)

= 574

At 95% confidence interval estimate of the population mean is,

- E < < + E

45420 - 574 < < 45420 + 574

44846 < < 45994

($44846 , $45994)

orchestra answered 1 year ago

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